Least-squares

In a least-squares, or linear regression, problem, we have measurements :math:A \in \mathcal{R}^{m \times n} and :math:b \in \mathcal{R}^m and seek a vector :math:x \in \mathcal{R}^{n} such that :math:Ax is close to :math:b. Closeness is defined as the sum of the squared differences:

.. math:: \sum_{i=1}^m (a_i^Tx - b_i)^2,

also known as the :math:\ell_2-norm squared, :math:\|Ax - b\|_2^2.

For example, we might have a dataset of :math:m users, each represented by :math:n features. Each row :math:a_i^T of :math:A is the features for user :math:i, while the corresponding entry :math:b_i of :math:b is the measurement we want to predict from :math:a_i^T, such as ad spending. The prediction is given by :math:a_i^Tx.

We find the optimal :math:x by solving the optimization problem

.. math::

   \begin{array}{ll}
   \mbox{minimize}   & \|Ax - b\|_2^2.
   \end{array}

Let :math:x^\star denote the optimal :math:x. The quantity :math:r = Ax^\star - b is known as the residual. If :math:\|r\|_2 = 0, we have a perfect fit.

Example

In the following code, we solve a least-squares problem with CVXPY.

.. code:: python

# Import packages.
import cvxpy as cp
import numpy as np

# Generate data.
m = 20
n = 15
np.random.seed(1)
A = np.random.randn(m, n)
b = np.random.randn(m)

# Define and solve the CVXPY problem.
x = cp.Variable(n)
cost = cp.sum_squares(A @ x - b)
prob = cp.Problem(cp.Minimize(cost))
prob.solve()

# Print result.
print("\nThe optimal value is", prob.value)
print("The optimal x is")
print(x.value)
print("The norm of the residual is ", cp.norm(A @ x - b, p=2).value)

.. parsed-literal::

The optimal value is 7.005909828287484
The optimal x is
[ 0.17492418 -0.38102551  0.34732251  0.0173098  -0.0845784  -0.08134019
  0.293119    0.27019762  0.17493179 -0.23953449  0.64097935 -0.41633637
  0.12799688  0.1063942  -0.32158411]
The norm of the residual is  2.6468679280023557